Benkelman beam overlay thickness numerical

Method

IRC:81 (Benkelman beam deflection method of overlay design), with the standard test load of 4085 kg on a dual wheel at 5.6 kg/cm2 tyre pressure. The question gives traffic only in CVPD with a growth rate, so the CVPD-based IRC:81 procedure (allowable deflection and $h=550\log_{10}(D_c/D_a)$) is used. The table gives 10 points (not 15), so 10 deflections are used.

Step 1 - Rebound deflection at each point

The dial readings are in units of 0.01 mm. Rebound deflection $=2\,(d_0-d_f)\times0.01$ mm (the factor 2 is the lever ratio of the beam). The question states that the lag corrections have already been made, so the intermediate readings are not used further and the final readings are taken.

Step 2 - Statistics

$\sum D=10.16$ mm, so the mean $\bar D=1.016$ mm. With $\sum(D-\bar D)^2=0.598$, the standard deviation is $\sigma=\sqrt{0.598/9}=0.258$ mm.

Step 3 - Characteristic deflection and corrections

• Characteristic deflection (IRC:81, general roads): $D=\bar D+\sigma=1.016+0.258=1.274$ mm.
• Temperature: the test was at 30 C and the standard is 35 C, so add 0.01 mm per degree: $1.274+0.05=1.324$ mm.
• Seasonal (subgrade moisture) factor 1.3: $D_c=1.324\times1.3=1.72$ mm.

Step 4 - Design traffic

$$A=P(1+r)^{n+10}=750\times(1.075)^{0+10}=1546\ \text{CVPD}$$

taking $n=0$ (overlay laid now); with $n=1$ or 2, $A=1662$ to 1786 CVPD. In every case the traffic is in the 1500-4500 CVPD class, for which the allowable deflection is $D_a=1.0$ mm (1.25 mm for 450-1500 CVPD, 1.5 mm for 150-450 CVPD).

Step 5 - Overlay thickness

Overlay thickness in bituminous macadam (BM):

$$h=550\log_{10}\frac{D_c}{D_a}=550\log_{10}\frac{1.72}{1.0}=550\times0.236=130\ \text{mm}$$

Answer: characteristic deflection $D_c\approx1.72$ mm; provide about 130 mm of BM overlay. Since 1 mm of BC/DBM is equivalent to about 1.5 mm of BM, this is about 90 mm of bituminous concrete/DBM (e.g. 50 mm DBM plus 40 mm BC).

Note: for a national or state highway, IRC:81-1997 uses $D_c=\bar D+2\sigma$. This gives $D_c=(1.016+0.516+0.05)\times1.3=2.06$ mm and $h=550\log_{10}(2.06/1.0)\approx172$ mm of BM. If the paper provides the IRC:81 overlay chart, read the thickness from it.